Find the equation of the straight lines passing through (1, 3) and (1) parallel to (2) perpendicular to the line passing through the point (3, 5) and (–6, 1).

Given A = (1, 3); B = (3, 5); C = (–6, 1)
$\text{ Slope of }\overline{\mathrm{BC}}=\frac{{\mathrm{y}}_2-{\mathrm{y}}_1}{{\mathrm{x}}_2-{\mathrm{x}}_1}$
$$\begin{gathered} m=\frac{1+5}{-6-3} \\ m=\frac{6}{-9} \\ m=\frac{-2}{3} \end{gathered}$$
(1) equation of line passing through A(1, 3) & $\text{ Parallel to }\overline{\mathrm{BC}}$ is
$\begin{array}{l}\mathrm{y}-{\mathrm{y}}_1=\mathrm{m}\left(\mathrm{x}-{\mathrm{x}}_1\right)\Rightarrow \mathrm{y}-3=\frac{-2}{3}(\mathrm{x}-1)\\ 3\mathrm{y}-9=-2\mathrm{x}+2\Rightarrow 3\mathrm{y}-9+2\mathrm{x}-2=0\\ 2\mathrm{x}+3\mathrm{y}-11=0\end{array}$
(2) equation of line passing through A(1, 3) $\text{ perpendicular to }\overline{\mathrm{BC}}$
$\begin{array}{l}{\perp }^{\mathrm{re}}\text{ slope }=\frac{-1}{\mathrm{m}}=\frac{-1}{\frac{-2}{3}}=\frac{3}{2}\\ \mathrm{y}-{\mathrm{y}}_1=\mathrm{m}\left(\mathrm{x}-{\mathrm{x}}_1\right)\Rightarrow \mathrm{y}-3=\frac{3}{2}(\mathrm{x}-1)\\ 2\mathrm{y}-6=3\mathrm{x}-3\Rightarrow 3\mathrm{x}-2\mathrm{y}+3=0\end{array}$